Answer:
I do not understand what you are asking
A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of flight, as it flies through x
Complete Question
The complete question is shown on the first uploaded image
Answer:
The temperature change is [tex]\frac{dT}{dt} = 1.016 ^oC/m[/tex]
Explanation:
From the question we are told that
The velocity field with which the bird is flying is [tex]\vec V = (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s[/tex]
The temperature of the room is [tex]T(x, y, u) = 400 -0.4y -0.6z-0.2(5 - x)^2 \ ^o C[/tex]
The time considered is t = 10 \ seconds
The distance that the bird flew is x = 1 m
Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird
Generally the change in the bird temperature with time is mathematically represented as
[tex]\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 * (5-x)] [-\frac{dx}{dt} ][/tex]
Here the negative sign in [tex]\frac{dx}{dt}[/tex] is because of the negative sign that is attached to x in the equation
So
[tex]\frac{dT}{dt} = -0.4v_y -0.6v_z -0.2[2 * (5-x)][ -v_x][/tex]
From the given equation of velocity field
[tex]v_x = 0.6x[/tex]
[tex]v_y = 0.2t[/tex]
[tex]v_z = -1.4 [/tex]
So
[tex]\frac{dT}{dt} = -0.4[0.2t] -0.6[-1.4] -0.2[2 * (5-x)][ -[0.6x]][/tex]
substituting the given values of x and t
[tex]\frac{dT}{dt} = -0.4[0.2(10)] -0.6[-1.4] -0.2[2 * (5-1)][ -[0.61]][/tex]
[tex]\frac{dT}{dt} = -0.8 +0.84 + 0.976[/tex]
[tex]\frac{dT}{dt} = 1.016 ^oC/m[/tex]
Luck walked to a store that is 250m away and it took him 50secs while Layne walked to the mall that is 1000m away and took her 200s to do. What do they have in common?
A. Average speed
B. Acceleration
C. Displacement
D.mass
Answer:
Average speed
Explanation:
250/50=5
1000/20=5
Which of the organisms in the food web above is the top level carnivore
Answer:
apex consumers
Explanation:
they are top
Which factor affects the amount of runoff that occurs in an area?
land use
the water table
the saturation zone
amount of nutrients in soil
Answer:
A. land use
Explanation:
Answer:
a
Explanation:
A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much work was done?
Answer:
The answer is 45 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distanceFrom the question
distance = 3 meters
force = 15 newtons
We have
workdone = 15 × 3
We have the final answer as
45 JHope this helps you
Pls help pls pls pls pls
am I right? be honest
Answer:
I chose c because it is the greater slope at point c
A
6. All other changeable factors that must
be kept the same to ensure a fair test
(what you keep the same).
Answer:
a constant variable?
Explanation:
A constant variable is any aspect of an experiment that a researcher intentionally keeps unchanged throughout an experiment.
Experiments are always testing for measurable change, which is the dependent variable. You can also think of a dependent variable as the result obtained from an experiment. It is dependent on the change that occurs
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.
Answer:
The number of turns in the solenoid is 22366.
Explanation:
The number of turns in the solenoid can be found using the following equation:
[tex] B = \mu_{0} I\frac{N}{L} [/tex]
Where:
B: is the magnetic field = 8.90 T
L: is the solenoid's length = 0.300 m
N: is the number of turns =?
I: is the current = 95 A
μ₀: is the magnetic constant = 4π×10⁻⁷ H/m
By solving equation (1) for N we have:
[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]
Therefore, the number of turns in the solenoid is 22366.
I hope it helps you!
Why wouldn't carbon dating work to determine the age of the earth?
A) Carbon dating works best on other planets
B) The half life of carbon is too short
C) The age of the earth cannot be determined
D) The half life of carbon is too long.
Answer:
The half-life of carbon is too short.
Explanation:
The answer is B.
Marisa’s car accelerates at an average rate of 2.6m/s^2. Calculate how long it takes her car to accelerate from 24.6m/s to 26.8m/s? Show your work.
given info is... Acceleration(a)=2.6m/s^2
final velocity(v)=26.8m/s
initial velocity(u)=24.6m/s
need to find.... time(t)=?
[tex]a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\[/tex]
[tex]t=\frac{v-u}{a}[/tex]
[tex]t=\frac{26.8-24.6}{2.6}[/tex]
[tex]t=0.846s[/tex]
Explanation:
It takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.
What is acceleration?Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.
Given parameters:
Initial speed of the car: u = 24.6 m/s
Final speed of the car: v = 26.8 m/s.
Acceleration of the car: a = 2.6 m/s²
Time interval: t = ?
change is speed = final speed - initial speed
= 26.8 m/s - 24.6 m/s
= 2.2 m/s
From the definition of acceleration,
acceleration = change is speed / time interval
So, time interval = change is speed / acceleration
= 2.2 m/s/2.6 m/s²
= 0.84 second.
Hence, it takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.
Learn more about acceleration here:
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Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.
The question is incomplete. Here is the complete question.
A current in the long, straight wire, which lies in the plane of rectangular loop, that also carries a current, as shown in the figure.
Find the magnitude of the net force exerted on the loop by the magnetic field created by the long wire. Answer in units of N.
Answer: Net Force = [tex]50.215.10^{-7}[/tex]N
Explanation: Force and Magnetic field are related through the following formula:
F = I.L.B.sinθ
Magnetic field (B) in a straight long wire is given by
[tex]B=\frac{\mu_{0}.I}{2.\pi.r}[/tex]
in which
[tex]\mu_{0}[/tex] is permeability of free space and is [tex]4.\pi.10^{-7}[/tex]T.m/A
I is current in the wire;
r is distance to the wire;
Examining the square loop and using the right hand rule, the top, which we will name it F₂, and the bottom, named F₄, have angle θ = 0, giving sin(0) = 0 and therefore, F₁ = F₃ = 0.
So, for the net force, the relevant forces will be on the sides parallel to the wire.
For the other forces, angle is 90°, sin(90°) = 1, then:
F = I.L.B
Replacing magnetic field:
F = [tex]\frac{\mu_{0}.I_{w}.L.I_{l}}{2.\pi.r}[/tex]
Note: The side closest to the wire is F₁, while the farthest is F₃.
Note2: As the constant unit is in meters, distance and length of side of the square loop are also in meters.
Calculating forces:
F₁ = [tex]\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.082}[/tex]
F₁ = [tex]278.975*10^{-7}[/tex]N
Current in F₃ is flowing thoruhg the negative side of the referential, so:
F₃ = [tex]-\frac{4*\pi*10^{-7}*4.3*0.19*14}{2.\pi.0.1}[/tex]
F₃ = [tex]-228.76*10^{-7}[/tex]N
Net force is total force:
[tex]F_{net} = F_{1}+F_{3}[/tex]
[tex]F_{net}=(278.975-228.76).10^{-7}[/tex]
[tex]F_{net}=50.22.10^{-7}[/tex]
The total force acting on the square loop is [tex]F_{net}=50.22.10^{-7}[/tex]N.
Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?
A) is the mechanical conserved between a and b? explain
B) is the mechanical energy conserved between b and c ?explain
Answer:
Yes at A the mechanical energy is conserved.
Yes at B the part of mechanical energy is conserved potential energy and kinetic energy and some is lost as frictional force.
Explanation:
Ratio = Eb/ Ea= 1058.3 J/2940 J= 0.3599
Ratio = Ec/ Eb= 0J/ 1058.3 J= 0
At point A the skater is at rest or it is the starting point and the whole energy is due to the position of the skater i.e= mgh = 50 *9.8*6= 2940 J
Since there's no movement there is no Kinetic energy = 0 J
Yes at A the mechanical energy is conserved.
At point B the skater has traveled for some of the distance . It has potential energy and kinetic energy.
Yes at B the part of mechanical energy is conserved as potential energy and kinetic energy.
The total Mechanical energy = 1058.3 J
At point B Total Mechanical energy = PE+ KE
1058.3J = 980 J + 78.3 J
1058.3 J = mgh + 1/2mv²
= 50*2*9.8 + 1/2 *50*(8.85)²
= 980 J + 78.3 J
As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .
2940 J-1058.3 J= 1881.7
At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME all are zero.
(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.
(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.
(c) mechanical energy is conserved between a and b.
(d) mechanical energy is not conserved between b and c.
The given parameters;
mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]The ratio of the mechanical energy at B and mechanical energy at A;
[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]
The ratio of the mechanical energy at C and mechanical energy at A;
[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]
Thus, we can conclude that mechanical energy is conserved between a and b.
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]
Thus, we can conclude that mechanical energy is not conserved between b and c.
Learn more here:https://brainly.com/question/19969393
How long is a day in Neptune
Answer: the long day in neptune would be .18383562 years!
Explanation:also for every day is 16 hours
The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
Answer:
The two charged objects will exert equal and opposite forces on each other.
Explanation:
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of charges on the objects and inversely proportional to the square of the distance between the two objects.
This force of attraction or repulsion between the two charged objects is always equal and opposite.
Therefore, the two charged objects will exert equal and opposite forces on each other.
g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
The average force exerted on the superball by the sidewalk is 0.00122 N.
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)
final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)
time of motion, t = 1800 s
The average force exerted on the superball by the sidewalk is given by;
[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]
Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.
A 849-kg car starts from rest on a horizontal roadway and accelerates eastward for 5.00 s when it reaches a speed of 35.0 m/s. What is the average force exerted on the car during this time
Answer:
The average force exerted on the car during this time is 5,943 N
Explanation:
Given;
mass of the car, m = 849 kg
initial velocity of the car, u = 0
time of motion of the car, t = 5.00 s
final velocity of the car, v = 35 m/s
The average force exerted on the car during this time is given by;
[tex]F = ma \\\\F = \frac{m(v-u)}{t}\\\\F = \frac{849(35-0)}{5}\\\\F = \frac{849(35)}{5}\\\\ F = 849*7\\\\F = 5,943 \ N[/tex]
Therefore, the average force exerted on the car during this time is 5,943 N
Answer:
5943N
Let's say (+x) = eastward
Average horizontal acceleration
ax = vx -v0x/5.00s
= 35.0m/s-0/5.00s
= +7.09m/s
From here we apply the second law of newton
During this period average horizontal force acting on car
Summation x = max = (849kg)(+7.09m/s²)
= 5943N
+5.943x10³N
= 5.94kN east ward.
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef
Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
g An angry rhino with a mass of 2700 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distraction, you throw a 0.180 kg rubber ball directly at the rhino with a speed of 9.05 m/s. Determine the speed of the ball (in m/s) after it bounces back elastically toward you.
Answer:
9.05m/s
Explanation:
given data
m1= 2700kg
v1=3.7m/s
m2=0.18kg
v2=9.05m/s
v3=?
We know that the velocity of the rhino will remains unchanged after impact as the mass of the rubber ball is negligible
m1v1+m2v2=m1v1+m2v3
2700*3.7+0.18*9.05=2700*3.7+0.18*v3
9990+1.629=9990+0.18v3
9991.629-9990=0.18v3
1.629=0.18v3
v3=1.629/0.18
v3=9.05m/s
How do I proton and and electron compared
On what part of the eye are rods and cones found?
Answer:
retina.
Explanation:
Two protons are a distance 3 10-9 m apart. What is the electric potential energy of the system consisting of the two protons
Answer:
The electric potential energy of the system is 7.87x10⁻²⁰ J.
Explanation:
The electric potential energy is given by:
[tex]E = \int{Fdr} = \frac{Kq_{1}q_{2}}{r}[/tex]
Where:
q₁ = q₂ is the charge of the protons = 1.62x10⁻¹⁹ C
r is the distance = 3x10⁻⁹ m
K: is the electrostatic constant = 9x10⁹ Nm²/C²
[tex] E = \frac{Kq_{1}q_{2}}{r} = \frac{9\cdot 10^{9} Nm^{2}/C^{2}*(1.62 \cdot 10^{-19} C)^{2}}{3\cdot 10^{-9} m} = 7.87 \cdot 10^{-20} J [/tex]
Therefore, the electric potential energy of the system is 7.87x10⁻²⁰ J.
I hope it helps you!
The electric potential energy of the system should be 7.87x10⁻²⁰ J.
Calculation of the electric potential energy:SInce We know that
fdr = kq1q2/r
Here
q₁ = q₂ i.e. is the charge of the protons = 1.62x10⁻¹⁹ C
r should be the distance = 3x10⁻⁹ m
K should be the electrostatic constant = 9x10⁹ Nm²/C²
Now electric potential energy should be
= (9x10⁹ Nm²/C² * 1.62x10⁻¹⁹ C) / 3x10⁻⁹ m
= 7.87x10⁻²⁰ J.
hence, The electric potential energy of the system should be 7.87x10⁻²⁰ J.
learn more about energy here: https://brainly.com/question/17384612
if you are driving 110 km/h along a straight road and you look to the side for 2.0 s , how far do you travel during this inattentive period ? explain.
Explanation:
hope this helps, have a good one :D
Answer:
60.12m
Explanation:
Distance = Velocity x Time
To use this formula we must first convert 110km/h to m/s, which we can do by dividing the value by 3.6:
110/3.6 = 30.56m/s (2dp)
Velocity = 30.56m/s
Time = 2s
Distance = 30.56x2
Distance = 61.12m
You travel 60.12m during this inattentive period.
Hope this helped!
How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R for resistance.)
Answer:
This relationship is explained by Ohm's law
Explanation:
Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as
V= IR
The bending of rocks due to the compression of tectonic plates is called
Ofaulting
O folding
subduction
plyometrics
Answer:
Folding
Explanation:
Which of these statements best describes the impact of ocean thermal power and current power on the environment?
A. Current power may decrease the fish population.
B. Current power may decrease the gravitational pull of the moon.
C. Ocean thermal power may increase the fish population.
C. Ocean thermal power may increase the gravitational pull of the moon.
Answer:
A. Current power may decrease the fish population.
Explanation:
The statement that best describes the impact of ocean thermal power and current power on the environment is that current power may decrease the fish population.
The environment is made up of living and non-living components that co-exist and interact with one another.
Harnessing current power from ocean movement will seriously affect the fish population. Most fishes are not sedentary. They move and glide through the water. When current power causes a change in the environment of the fish. This will definitely affect the normal condition prevalent in the body of water.Answer:
A
Explanation:
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A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]
Explanation:
From the question we are told that
The weight of the mountain climber is m = 555 N
Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as
[tex]T_a* cos 65 -555 + T_b * cos(85) = 0[/tex]
Here [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side
So
[tex]0.423T_a + 0.0871T_b = 555[/tex]
=> [tex] 0.0871T_b = 555 - 0.423T_a[/tex]
=> [tex] T_b = \frac{555 - 0.423T_a}{0.0871}[/tex]
Generally from the diagram , the total amount of force acting on the rope along the horizontal axis at equilibrium is mathematically represented as
[tex]T_a* sin 65 - T_b * sin(85) = 0[/tex]
=> [tex] 0.9063T_a - 0.9962T_b = 0[/tex]
=> [tex] 0.9063T_a = 0.9962T_b [/tex]
=> [tex] 0.9063T_a = 0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]
=> [tex] 0.9063T_a = [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]
=> [tex] 0.0789T_a = [552.891 - 0.421T_a[/tex]
=> [tex] 0.4999T_a = 552.891 [/tex]
=> [tex] T_a = 1106 \ N [/tex]
12, The error in measurement may occur due to
S
a, inexperience of a person
b, The faulty apparatus
c, Inappropriate method
d, Due to all reasons in a, b and c
Answer:
d, Due to all reasons in a, b and c
Explanation:
All of the reasons from the choices can lead to errors in measurement. An error is an uncertainty introduced to a scientific process.
Errors can be due to the following reasons;
It can be to a poor technician; an inexperienced scientist can introduce serious into their set up. Even when recording their observation, a dearth of experience can lead to error. Faulty apparatus can also lead to errors in measurement. An instrument that is poorly calibrated can also result in measurement errors. An inappropriate scientific method or measuring guidelines can also lead to errors in measurements.block of mass m sits at rest on a rough inclined ramp that makes an angle with the horizontal. What must be true about normal force F on the block due to the ramp
Answer:
Explanation:
For a body on a ramp with mass m, the forces acting on the body along the vertical component are the weight and the normal reaction.
The weight of the body acts in the negative y direction while the normal reaction acts in the positive y direction
Taking the sum of forces along the y component
Sum Fy = -W+R = ma
Since acceleration is zero
-W+R = m(0)
-W+R = 0
-W = -R
W = R
Hence the Normal reaction force acting on the on the body is equal to normal force